Part A Part B Suppose that you have a reflection diffraction grating with n= 750

Question

Part A Part B Suppose that you have a reflection diffraction grating with n= 750 lines per millimeter. Lightfrom a sodium lamp passes through the grating and is diffractedonto a distant screen Part A Two visible lines in the sodium spectrum havewavelengths 498 nm and 569 nm. What is the angular separationDelta theta of the first maxima ofthese spectral lines generated by this diffraction grating? Express your answer in degrees to two significant figures Delta theta =3.3 degree Part B How wide does this grating need to be toallow you to resolve the two lines 589.00 and 589.59 nanometers,which are a well known pair of lines for sodium, in the secondorder (m=2)? Express your answer in millimeters totwo significant figures. ___________________ mm?

Explanation / Answer

Question is a bit odd - if it's a reflection grating, then lightdoesn't pass through it... Also, the angle of incidence is not given, but let's assume that's0 deg (i.e. normal). We know that the phase varies with angle according to=2/*bsin. So to find out how phase varieswith angle when you change the angle very slightly, wedifferentiate: d=2/*bcosd. Now the very small angle is going to be the angle of the next min(re: Rayleigh criterion). For N slit diffraction, the change inphase btw max and the next min is d=2/N, but we nowknow how to turn a change in phase into a change in angle bysubstituting in above expression for d: bcosd=/N   - which tells us how anglebtw max and min changes with number of slits. Specifically at a big maximum, =2m som=bsin of course. How does this position change with asmall change in wavelength? We differentiate again: md=bcosd but we already know how angle varies with N so we can substitutethat in to find how wavelength varies with N: md=/N So our wavelength resolution /d=mN If we need to resolve 589.00 and 589.59nm in 2nd order (for someobscure reason), the mean =589.295nm and d=0.590nm,m=2 N=499.4 (but N has to be an integer so N=500). since the spacing between the lines is b=1e-3/750=1.33e-6m, thenthe width of 500 lines is 500*1.33e-6m (or 499*1.33e-6m if you wantto be pedantic) = 0.67mm (better check my calcs!)

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